# The position of an object moving along a line is given by p(t) = t - 3sin(( pi )/3t) . What is the speed of the object at t = 4 ?

Nov 26, 2017

$p \left(t\right) = t - 3 \sin \left(\frac{\pi}{3} t\right)$
$t = 0 \implies p \left(0\right) = 0 m$
$t = 4 \implies p \left(4\right) = 4 - 3 \sin \left(\frac{\pi}{3} \cdot 4\right) \implies$
$p \left(4\right) = 4 - 3 \sin \left(\pi + \frac{\pi}{3}\right)$ (1)
$\sin \left(\pi + t\right) = - \sin \left(t\right)$ (2)
(1)+(2)$\implies$$p \left(4\right) = 4 - \left(3 \cdot \left(-\right) \sin \left(\frac{\pi}{3}\right)\right) \implies$
$p \left(4\right) = 4 + 3 \cdot \frac{\sqrt{3}}{2}$
$p \left(4\right) = \frac{8 + 3 \sqrt{3}}{2} m$

Now it depends on the extra information given:

1.If the acceleration isn't constant:
Using the law of space for the varied linear uniform movement:
d=V""_0*t+(a*t^2)/2
where
$d$ is the distance,V""_0 is the initial speed,$a$ is the acceleration and $t$ is the time when the object is in position $d$.

$p \left(4\right) - p \left(0\right) = d$
Assuming that the initial speed of the object is $0 \frac{m}{s}$
$\frac{8 + 3 \sqrt{3}}{2} = 0 \cdot 4 + \frac{a \cdot 16}{2} \implies$
$a = \frac{8 + 3 \sqrt{3}}{16} \frac{m}{s} ^ 2$

Finally the speed of the object at t=4 is
$V = a \cdot 4 = \frac{8 + 3 \sqrt{3}}{4} \frac{m}{s}$

2.If the acceleration is constant:
With the law of linear uniform movement:
p(4)=p(0)+V(t-t""_0)
You will get:
$\frac{8 + 3 \sqrt{3}}{2} = 0 + V \cdot 4 \implies$
$V = \frac{8 + 3 \sqrt{3}}{8} \frac{m}{s}$